RD Sharma Solutions Class 7 Congruence Exercise 16.2

RD Sharma Solutions Class 7 Chapter 16 Exercise 16.2

RD Sharma Class 7 Solutions Chapter 16 Ex 16.2 PDF Free Download

Exercise 16.2

Q1. In the following pairs of triangle (Fig. 12 to 15), the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic.

1
2

Answer:

3

1) In \(\Delta\) ABC and \(\Delta\) DEF

AB = DE = 4.5 cm (Side)

BC = EF = 6 cm (Side) and

AC = DF = 4 cm (Side)

Therefore, by SSS criterion of congruence, \(\Delta ABC \cong \Delta DEF\)

2)

4

In \(\Delta\) ACB and \(\Delta\) ADB

AC = AD (Side)

BC = BD (Side) and

AB = AB (Side)

Therefore, by SSS criterion of congruence, \(\Delta ACB \cong \Delta ADB\)

3) In \(\Delta\) ABD and \(\Delta\) FEC,

AB = FE (Side)

AD = FC (Side)

BD = CE (Side)

Therefore, by SSS criterion of congruence, \(\Delta ABD \cong \Delta FEC\)

4) In \(\Delta\) ABO and \(\Delta\) DOC,

AB = DC (Side)

AO = OC (Side)

BO = OD (Side)

Therefore, by SSS criterion of congruence, \(\Delta ABO \cong \Delta ODC\)

Q2. In figure, AD = DC and AB = BC

(i) Is \(\Delta ABD \cong \Delta CBD\)?

(ii) State the three parts of matching pairs you have used to answer (i).

7

Answer:

Yes \(\Delta\) ABD = \(\Delta\) CBD by the SSS criterion. We have used the three conditions in the SSS criterion as follows:

AD = DC

AB = BC and

DB = BD

 

Q3. In Figure, AB = DC and BC = AD.

(i) Is \(\Delta ABC \cong \Delta CDA\)?

(ii) What congruence condition have you used?

(iii) You have used some fact, not given in the question, what is that?

8

Answer:

We have AB = DC

BC = AD

and AC = AC

Therefore by SSS \(\Delta ABC \cong \Delta CDA\)

We have used Side Side Side congruence condition with one side common in both the triangles.

Yes, have used the fact that AC = CA.

 

Q4. In \(\Delta PQR \cong \Delta EFD\),

(i) Which side of \(\Delta PQR\) equals ED?

(ii) Which angle of \(\Delta PQR\) equals angle E?

Answer:

\(\Delta PQR \cong \Delta EFD\)

(i) Therefore PR = ED since the corresponding sides of congruent triangles are equal.

(ii) \(\angle\) QPR = \(\angle\) FED since the corresponding angles of congruent triangles are equal.

9

Q5. Triangles ABC and PQR are both isosceles with AB = AC and PO = PR respectively. If also, AB = PQ and BC = QR, are the two triangles congruent? Which condition do you use?

It \(\angle\) B= 50°, what is the measure of \(\angle\) R?

Answer:

We have AB = AC in isosceles \(\Delta\) ABC

And PQ = PR in isosceles \(\Delta\) PQR.

Also, we are given that AB = PQ and QR = BC.

Therefore, AC = PR (AB = AC, PQ = PR and AB = PQ)

Hence, \(\Delta ABC \cong \Delta PQR\)

Now

\(\angle\) ABC = \(\angle\) PQR (Since triangles are congruent)However, \(\Delta\)  PQR is isosceles.

Therefore, \(\angle\) PRQ = \(\angle\) PQR =  \(\angle\) ABC = 50°

 

Q6. ABC and DBC are both isosceles triangles on a common base BC such that A and D lie on the same side of BC. Are triangles ADB and ADC congruent? Which condition do you use? If  \(\angle\) BAC= 40° and \(\angle\) BDC = 100°, then find \(\angle\) ADB.

Answer,

Yes, triangles ADB and ADC are congruent (By SSS)

AB=AC, DB = DC and AD = DA

\(\angle\) BAD = \(\angle\) CAD (c.p.c.t)

\(\angle\) BAD + \(\angle\) CAD = 40°/ 2 \(\angle\) BAD = 40°

\(\angle\) BAD = 40°/2 =20°

\(\angle\) ABC + \(\angle\) BCA + \(\angle\) BAC = 180° (Angle sum property)

Since \(\Delta\) ABC is an isosceles triangle,

\(\angle\) ABC = \(\angle\) BCA \(\angle\) ABC +\(\angle\) ABC + 40°= 180°

2 \(\angle\) ABC = 180° – 40° = 140° \(\angle\) ABC = 140°/2 = 70°

\(\angle\) DBC + \(\angle\)  BCD + \(\angle\)  BDC = 180° (Angle sum property)

Since \(\Delta\) ABC is an isosceles triangle, \(\angle\)  DBC = \(\angle\) BCD \(\angle\) DBC + \(\angle\) DBC + 100o = 180°

2 \(\angle\) DBC = 180° – 100o = 80°

\(\angle\) DBC=80°/2=40°

In \(\Delta\) BAD,

\(\angle\) ABD + \(\angle\) BAD + \(\angle\) ADB = 180°(Angle sum property)

30° + 20° + \(\angle\) ADB = 180° (\(\angle\) ADB = \(\angle\) ABC – \(\angle\) DBC), \(\angle\) ADB =180°- 20° – 30°

\(\angle\) ADB = 130°

\(\angle\) ADB =130°

 

Q7. \(\Delta\) ABC and \(\Delta\) ABD are on a common base AB, and AC = BD and BC = AD as shown in Fig. 18. Which of the following statements is true?

(i) \(\Delta ABC \cong \Delta ABD\)

(ii) \(\Delta ABC \cong \Delta ADB\)

(iii) \(\Delta ABC \cong \Delta BAD\)

10

Answer:

In \(\Delta\) ABC and \(\Delta\) BAD we have,

AC = BD (given)

BC = AD (given)

and AB = BA (common)

Therefore by SSS criterion of congruency,  \(\Delta ABC \cong \Delta BAD\)

There option (iii) is true.

 

Q8. In Figure, \(\Delta\) ABC is isosceles with AB = AC, D is the mid-point of  base BC.

(i) Is \(\Delta ADB \cong \Delta ADC\)?  

(ii) State the three pairs of matching parts you use to arrive at your answer.

11

Answer:

We have AB = AC.

Also since D is the midpoint of BC, BD = DC

Also, AD = DA

Therefore by SSS condition,

\(\Delta ADB \cong \Delta ADC\)

We have used AB, AC : BD, DC AND AD, DA

 

Q9. In figure, \(\Delta\) ABC is isosceles with AB = AC. State if \(\Delta ABC \cong \Delta ACB\). If yes, state three relations that you use to arrive at your answer.

12

Answer:

Yes, \(\Delta ABC \cong \Delta ACB\) by SSS condition.

Since, ABC is an isosceles triangle, AB = BC, BC = CB and AC = AB

 

Q10. Triangles ABC and DBC have side BC common, AB = BD and AC = CD. Are the two triangles congruent? State in symbolic form, which congruence do you use? Does \(\angle\) ABD equal \(\angle\) ACD? Why or why not?

Answer:

Yes,

Given,

\(\Delta\) ABC and \(\Delta\) DBC have side BC common, AB = BD and AC = CD

By SSS criterion of congruency, \(\Delta ABC \cong \Delta DBC\)

No, \(\angle\) ABD and \(\angle\) ACD are not equal because AB \(\neq\) AC

13


Practise This Question

Two players A and B throw a die alternately for a prize of Rs 11, which is to be won by a player who first throws a six.  If A starts the game, their respective expectations are